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Titrate sulphuirc acid with sodium hydroxide 
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#1
Mar605, 10:57 AM

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Coursework time has hit me at college and no matter how much I hit the books the answer is not hitting me.
It might be simple but I can't see it. Of course, the question is needed. I had to titrate sulphuirc acid with sodium hydroxide. Not hard to write coursework on and I am sure that if I did it without what I am trying to find out (below) then it would be fine but I am not happy just to accept that there should be equal amounts of sulphuric acid and sodium hydroxide to create a neutral solution. I read in these forums that the molar pH of sulphuirc acid is less than that of the pOH of the sodium hydroxide. For my hypothesis I want to be able to say why I think that there needs to be more sodium hydroixde to create a neutral pH overall. For what I need help with is the maths, really. I have been studying for the last 4 hours trying to make sense of the dissonance(?) of acids and bases and how this affects the pH or the moles (not even sure which). I installed a system that was posted up here from the 'few questions' thread (called BATE) and it says that for an equal amount of [tex]0.01[/tex] [tex]mol[/tex] [tex]dm^{3}[/tex] concentrations of each susbstance I should get an overall pH of 6.17. I am trying to find out how to get to this number and how I can explain it along the way. Even if this wasn't coursework I would have ask the same question (actually, to be honest I don't need it in my coursework it is more for interest but I really want to understand it). Cheers The Bob (2004 ©) P.S. After rereading it I hope it makes sense. What I want to know is how to work out that the pH after mixing, say, 10cm³ of sodium hydroixde to 10cm³ of sulphuric acid (both concentrated to 0.01M) is going to be 6.17, and therefore how much sodium hydroxide is actually needed. Also I will need to explain why so if you know any websites that will do that, to save you writing it, then I would be grateful. 


#2
Mar605, 04:27 PM

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The pH of an equimolar sodium hydroxide and sulfuric acid is 7, both are considered strong acid/base; the product salt of the two is weak with respect acid or base and thus the pH will remain 7. In the case of a different combination of where one of them is a weak acid/base and the other a strong base/acid with equimolar values of both, you'll need to consider that the product salt component of the weak acid/base will actually influence the pH upon the complete neutralization.
Yes, the strong base acid/base will react to completion with the weak acid/base (that is completely neutralized)...the sole factor which affects the pH afterwards is the product salt. 


#3
Mar605, 04:41 PM

P: 1,116

I have done the titration and the results show that for 10.0cm³ of 0.01M sulphuric acid, approximately 11.0cm³ of 0.01M sodium hydroixde is needed. Are you trying to say that on of the substances is strong and the other is weak because with equal concentrations I can't see that making any difference.
I have been looking around and I found these sorts of equations: [tex]K_w = [H^+][OH^][/tex] [tex]K_a = \frac{[H^+][HA^]}{[H_2 A]}[/tex] coming from [tex]H_2A \rightarrow H^+ + HA^[/tex] The Bob (2004 ©) 


#4
Mar605, 05:58 PM

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Titrate sulphuirc acid with sodium hydroxide
Chemical calculators for labs and education BATE  pH calculations, titration curves, hydrolisis 


#5
Mar605, 06:04 PM

Admin
P: 23,397

Chemical calculators for labs and education BATE  pH calculations, titration curves, hydrolisis 


#6
Mar705, 09:50 AM

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If you're referring to to the second ionization of sulfuric acid, it is negligible compared to the first. Look up the ionization constants yourself. 


#7
Mar705, 10:00 AM

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PF Gold
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The pH comes from the concentration of the acid/base as well as the dissociation constant (or you can use Ka/Kb). If the dissociation constants are very close to 1 (Ka/Kb large), the pH is roughly a function of concentration alone.



#8
Mar705, 10:29 AM

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Chemical calculators for labs and education BATE  pH calculations, titration curves, hydrolisis 


#9
Mar705, 11:38 AM

P: 1,116

Cheers. The Bob (2004 ©) 


#10
Mar705, 02:12 PM

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Bob, solve this equation for x, assuming a original solution of sulfuric acid of .01M and sodium hydroxide .01M... [tex].012= \frac{[x]^2}{[.01Mx]}[/tex] Subtract from pH 7. What you can also do is to actually calculate the pH of sulfuric acid as a whole, including Ka for both its acids. After finding the total concentration of H+ produced, find the excess of the latter by subtracting from the concentration of hydroxide. This would be more accurate. [tex].012= \frac{[.01+x][x]}{[.01x]} [/tex], solve for x. [itex][H_3O^+]=[x+.01][/itex]. I'm sure you can do the rest. You'll find that the pH of an equimolar solution of sodium hydroxide and sulfuric acid is roughtly 6.9, perhaps a bit higher. 


#11
Mar705, 05:04 PM

P: 1,116

So how do you know that [tex]K _{a2}[/tex] is 0.012 and why is it that you are using [tex]K _{a2}[/tex]??? The Bob (2004 ©) 


#12
Mar705, 05:14 PM

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Chemical calculators for labs and education BATE  pH calculations, titration curves, hydrolisis 


#13
Mar705, 05:18 PM

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Here is the problem. I understand nothing at all to do with working out pHs or pOHs. I only found out yesterday that there was a unit pOH.
The problem is I have 10 cm^{3} of Sodium Hydroxide and 10 cm^{3} of Sulphuric Acid. I mix them together and they, obviously, create an overall pH. I need to work out what that pH is. How do I do it, using the formulas that are needed and an explaination to what each bit is for and means, please. I would be really, really grateful if someone could explain it to be and, more or less, drag me through it. Thanks. The Bob (2004 ©) 


#14
Mar705, 06:12 PM

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Dissociation equation: [tex]AcH \leftrightarrow Ac^ + H^+[/tex] Equilibrium constant: [tex]K_a = \frac{[Ac^][H^+]}{[AcH]}[/tex] If you know total concentration  say it is [tex]C_a[/tex]  you may calculate [tex]H^+[/tex] concentration assuming  in accordance with reaction equation  that [tex][Ac^] = [H^+][/tex]. If so, undissociated acid concentration is [tex][AcH] = C_a  [H^+][/tex] and [tex]K_a = \frac{[H^+]^2}{C_a  [H^+]}[/tex] which is a quadratic equation. Solve it for [tex][H^+][/tex]: [tex][H^+] = \frac{K_a + \sqrt{K_a^2 + 4 K_a C_a}}{2}[/tex] and you know how to perform pH calculation of the weak acid solution. Assumption that [tex][Ac^] = [H^+][/tex] is not always valid  to be precise you should take account of [tex][H^+][/tex] ions from the water autodissociation. For not very weak acids and not very diluted solutions assumption holds. Now get back to the sulphuric acid and its equimolar solution with sodium hydroxide  which is the same as [tex]NaHSO_4[/tex] solution. You may treat [tex]HSO_4^[/tex] the same way acetic acid was treated above  first dissociation step was already 'consumed' by neutralization and is not influencing the situation (that's not always the case  here is). As I already wrote [tex]pK_{a2}[/tex] equals 2  so [tex]K_{a2}[/tex] (which describes second dissociation equilibrium) equals 0.01. Put it into the equation for the [tex][H^+][/tex] together with the 0.01 concentration (in the real titration you will have to calculate the dilution factor) and you will get [tex][H^+] = \frac{0.01 + \sqrt{0.01^2 + 4*0.01*0.01}}{2} = 0.00618[/tex] and pH = 2.21 Hope that helps. Chemical calculators for labs and education BATE  pH calculations, titration curves, hydrolisis 


#15
Mar705, 06:23 PM

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Borek, I mentioned a solution of the strong acid base, however you were implying that I had made a statement where an pure acid solution had a pH of 7.
Bob, I'm not able to explain everything about acid/base chemistry right here, what you'll need to do is read up on your own. For now I'll explain a simple case...what is the pH of a solution of .01M sulfuric acid? We'll assume that the first acid dissociates to completion, this will give .01M of [H+] and [HSO4] upon dissociation. The pH at this point can be calculated by taking the negative log of the concentration of [H+]....[itex]pH=log[H^+][/itex], at this point the pH is 2. However, if you wish to be absolutely exact the you'll need to incorporate the second ionization which has its own dissociation constant [itex]K_{a2}[/itex]. This is where this equation comes in [tex]K_{a2}= \frac{[H+][SO_4^{2}]}{[HSO_4^]} [/tex] the initial concentration of [H+] is .01M, 0 for the sulfate, and .01 for the acid. Thus when x M of the acid has dissociated, [tex].012= \frac{[.01+x][x]}{[.01x]} [/tex] Solving for x and then finding [.01 + x] will give you the hydronium concentration of which the pH is approximately 1.9. 


#16
Mar805, 03:29 PM

P: 1,116

The rest of this post I can, again, accept, so long as if I try another one I can run it through here. Cheers for the help so far. The Bob (2004 ©) 


#17
Mar805, 03:34 PM

P: 1,116

The Bob (2004 ©) 


#18
Mar805, 04:46 PM

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It is Ka2, the equilibrium constant for the second dissociation of sulfuric acid. So again, the summary is that by neglecting this dissociation, a sulfuric acid solution will be deemed to have a pH of 2; by taking account of the second dissociation, it will roughly have a pH of 1.9. From this you can obtain a good idea of the pH due to equimolar solution pertaining to your initial proposal. For most calculations at most schools/universities, the second dissociation is neglected, so you won't have an opportunity to show off your skills.



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